AP ICET 2015 Paper

Given digits : 0,1,2,3,4
4-digit number :

I

II

III

IV

Total no. of 4–digit numbers (Repetition of digits not allowed ) : I can be filled in 4 ways as it can’t have zero, II can be filled in 4 ways by the remaining
4 digits, III in 3 ways and IV in 2 ways
∴ Total no. of outcomes $=4\times 4\times 3\times 2=96$
Similarly,
No. of 4-digit numbers ending with 0 $=4\times 3\times 2=24$
No. of 4-digit numbers ending with 2$=3\times 3\times 2=18$
No. of 4-digit numbers ending with 4$=3\times 3\times 2=18$
No. of favourable outcomes $=24+18+18=60$
Required probability $=\frac{60}{96}=\frac{5}{8}$